anonymous
  • anonymous
PLEASE HELP!!!!!!!! what is log base 3 six square root 243 (where 6 is inside the "v" of the square root"
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\log _{3} \sqrt[6]{243}\]
anonymous
  • anonymous
\[\log_{3}\sqrt[6]{243} \] like this?
anonymous
  • anonymous
yes.

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anonymous
  • anonymous
alright lemme see what i can do
anonymous
  • anonymous
okay, can u help me with a few others as well? i am really not understanding logs :(
anonymous
  • anonymous
no problem. You know the rule of logs? Where logANS / logBASE = expo?
anonymous
  • anonymous
?
anonymous
  • anonymous
so in this case log [\sqrt[6]{243}\] / log3 = 5/6
anonymous
  • anonymous
i don't really get it..can you draw it out?
anonymous
  • anonymous
|dw:1327550139579:dw|
anonymous
  • anonymous
iam not allowed to use a calculator for these so i have to get it in the exponent form
anonymous
  • anonymous
\[\frac{1}{6}\log_3(243)\] is a start
anonymous
  • anonymous
and since \[243=3^5\] you get \[\frac{1}{6}\log_3(3^5)=\frac{5}{6}\]
anonymous
  • anonymous
|dw:1327550493647:dw|
anonymous
  • anonymous
was i on the right rack?
anonymous
  • anonymous
Yes. Now use logs to make the expo turn into a coefficient
anonymous
  • anonymous
how do i do that ...?
anonymous
  • anonymous
log3^y = log243^(1/6)
anonymous
  • anonymous
Now you can put the y in front of the log as well as the 1/6 so it becomes ylog3 = (1/6)[log243]
anonymous
  • anonymous
i am so confused :'''(
anonymous
  • anonymous
shall we go off satelite73's method? We keep it as a log form.
anonymous
  • anonymous
i have to do this non calc
anonymous
  • anonymous
So from the \[\sqrt[6]{}\] we can turn it into an exponent as 1/6 so\[\log_{3}243^{1/6}\]
anonymous
  • anonymous
i get that now, so im using satelites method as that does not require a calc.
anonymous
  • anonymous
really your life would be a lot easier if you put the 1/6 out front
anonymous
  • anonymous
i said that too
anonymous
  • anonymous
\[\log(x^n)=n\log(x)\] so \[\log(\sqrt[6]{x})=\frac{1}{6}\log(x)\]
anonymous
  • anonymous
i am really not understanding anything. i just can't do this :(
anonymous
  • anonymous
The reason we use logrithims for unknown exponential equations is to turn it from an exponent to a coefficient. It then becomes an algebraic equation. So in this situation we have \[\log_{3}\sqrt[6]{243} \] We turn the \[\sqrt[6]{243}\] into \[243^{1/6}\] So right now we have \[\log_{3} 243^{1/6}\] now since we have the log, we are allowed to drop the 1/6 down from the exponent level to become the leading coefficient thus making the equation \[1/6\log_{3}243 \] From here we can change the number 243 into a number with a base of 3. We know that 3^5 = 243 therefore we can write \[1/6\log_{3}(3^5) \] And knowing that when a base and the number above it are the same number u can cancel out the log and it's base, leaving you with just the 5. So you are to be left with 1/6 * 5 this is known to be 5/6
anonymous
  • anonymous
dang. thank you so much! alb to help me with anymore?
anonymous
  • anonymous
Yessir! glad to help :)

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